\( 26 \% \) of college students say they use credit cards because of the rewards program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the rewards program is (a) exactly two, (b) more than two, and (c) between two and five inclusive. If convenient, use technology to find the probabilities. (a) \( \mathrm{P}(2)=0.274 \) (Round to the nearest thousandth as needed.) (b) \( \mathrm{P}(\mathrm{x}>2)=0.503 \) (Round to the nearest thousandth as needed.) (c) \( \mathrm{P}(2 \leq x \leq 5)=\square \) (Round to the nearest thousandth as needed.)

To solve this problem, we can model the situation using the binomial probability formula since we have a fixed number of trials (10 college students), two possible outcomes (using credit cards for rewards or not), and a constant probability of success (26% or 0.26). First, let’s note the formula for finding the probability of exactly \( k \) successes in \( n \) trials: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where: - \( \binom{n}{k} \) is the binomial coefficient, - \( n \) is the number of trials, - \( k \) is the number of successes, - \( p \) is the probability of success, and - \( (1-p) \) is the probability of failure. Given: - \( n = 10 \) - \( p = 0.26 \) (a) For \( P(X = 2) \): \[ P(X = 2) = \binom{10}{2} (0.26)^2 (0.74)^{8} \] Calculating this we get: \[ = \frac{10!}{2!(10-2)!} \cdot (0.26)^2 \cdot (0.74)^{8} \approx 0.274 \] (b) For \( P(X > 2) \): This can be computed as \( 1 - P(X \leq 2) \) which is \( 1 - [P(X = 0) + P(X = 1) + P(X = 2)] \). Calculating each separately: - \( P(X = 0) \approx 0.062 \) - \( P(X = 1) \approx 0.187 \) - \( P(X = 2) \approx 0.274 \) Thus, \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.062 + 0.187 + 0.274 \approx 0.523 \] So, \[ P(X > 2) \approx 1 - 0.523 \approx 0.477 \] (c) For \( P(2 \leq X \leq 5) \): This is calculated as \( P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \). Calculating \( P(X = 3), P(X = 4), \) and \( P(X = 5) \): Using the formula: - \( P(X = 3) \approx 0.227 \) - \( P(X = 4) \approx 0.137 \) - \( P(X = 5) \approx 0.062 \) Adding them together: \[ P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \approx 0.274 + 0.227 + 0.137 + 0.062 \approx 0.700 \] So the final results (rounded to the nearest thousandth) are: **(a)** \( P(2) = 0.274 \) **(b)** \( P(X > 2) = 0.477 \) (Corrected from earlier, make sure to feed this into your tech tools for verification!) **(c)** \( P(2 \leq X \leq 5) \approx 0.700 \) (Round to the nearest thousandth as needed.)